Merten's theorem on cauchy products

Question

Suppose we know that $$\{a_n\}$$ is a sequence such that $$\sum_{n=0}^{\infty}a_n=0$$ and that for some $N$ quite large, we have that $a_n=0$ for any $n\geq N$. Then we notice that the infinite series of $a_n$ converve absolutely. If we were to now compute the cauchy product of $a_n$ with some $b_n$, we would find $$\sum_{n=0}^{\infty}\sum_{k=0}^na_kb_{n-k}=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}a_kb_{n-k}=\sum_{k=0}^{\infty}a_k\sum_{n=k}^{\infty}b_{n-k}$$ Thus, if we can find a sequence $\{b_n\}$ such that $\sum_{n=k}^{\infty}b_{n-k}$ converges to $-k$, then we will have that this cauchy product $$=\sum_{k=0}^{\infty}a_k(-k)$$ and that this is equal to $0$ by mertens' theorem. So, my question is twofold: can anyone find an appropriate sequence $b_n$? and/or can anyone find a mistake in the above reasoning. Please note that something like $k\sum$(of something that adds to 1) as $b_n$ won't work with Mertens' theorem.

Answer 1

It seems to me that $$\sum_{n=k}^\infty b_{n-k}=-k$$ is the same as $$\sum_{n=0}^\infty b_{n}=-k,$$ and it does not make a whole lot of sense to expect that to hold for all $k$.